# Simple harmonic motion confusion

While doing maths for SHM in terms of circular motion, say an object is rotating in anticlockwise direction. It moves from point A to point B,both points being located in radius of the circle (imagined). Then why do we take the displacement as the perpendicular line drawn from the point B to the radius at starting position?Even if we say the horizontal distance traveled do not help to increase the acceleration of the body(given by a∝ya∝y), we still use yy to calculate the velocity of the body where yy is not even true displacement but just a perpendicular drawn to complete the right angle?
Sorry if the question doesn't even make sense,I'm don't have good understanding of physics.

• The circle in this exercise (sometimes called the "reference circle") is just a computational aid. No physical object moves in a circle when a system undergoes SHM unless a teacher goes to great trouble to set up a parallel pair of systems to show the relationship. You can dispense with the reference circle entirely if you can solve the differential equation of motion directly. But if you *can't (yet) solve differential equations you can notice that the projection of uniform circular motion meets the requirements for SHM (acceleration and velocity at the center and the limits of motion). – dmckee Dec 30 '16 at 15:51
• physics.stackexchange.com/a/398055/150025 might help you out. – Yuzuriha Inori Apr 10 '18 at 2:15
• Please, consider editing your question to make it clearer. Can you separate it in paragraphs? Could you please upload a picture of what you mean? It doesn't have to be perfect, just a sketch, but it'd be really helpful. – FGSUZ Jul 24 '18 at 0:57
• What you do exactly is constructing a phasor diagram to solve the problem – Aditya Garg Nov 11 '18 at 22:19

One dimensional SHM $$x(t)=k\sin (\omega t)$$ can be represented in two-dimensional phase space where the co-ordinates are $$x(t)$$ and $$\dot x(t)$$. In phase space we have

$$(x(t), \dot x(t)) = (k \sin (\omega t), k \omega \cos (\omega t))$$

which is the parametric equation of an ellipse. With appropriate scaling of the axes, this ellipse becomes a circle - so SHM is often represented as a circle in phase space.

As has been said in other answers, nothing actually moves in a circle in physical space, and the velocity of the particle is simply the second co-ordinate $$\dot x(t)$$ in phase space, which is usually plotted vertically in a phase space diagram.

Start with a spring, a real physical spring, that just bounces up and down. Now you need to draw this motion on a piece of paper.

I don't think you have any trouble drawing the mass and the spring moving in and out from the origin to an X coordinate and back again.

This same idea of not having to duplicate on paper exactly what is actually happening physically is why we can draw simple harmonic motion in a circle.

The main point of s.h.m , the in and out motion on your drawing and the drawing of a circle is that they all convert a repetitive motion, that travels the same distance every time, into a mental picture of the basic idea.

That all you want to do really, if you think about it, even without drawing a picture, the use of the cos and sin functions in the equation, because they repeat, conveys the same idea.

• $\omega$ is angular velocity. You probably know know, but the circle, represents a complete shm cycle. I think the YouTube videos, if you can find a good one, are the best way to learn if you are having trouble with the way they draw things out. You might be getting confused with the different ways its drawn, but its all the same idea. – user140606 Dec 30 '16 at 4:44